anonymous
  • anonymous
solving rational equations: x+2 x - 1 4x+2 ---- = ---- - ----- x x x^2-3x I know how to make them all have a common denominator x(x-3) but when it comes to solving I get x^2+5x+6=x^2-4x-3x-3x-2 which then comes up to be x^2+3x-11 This isn't right. What did I do wrong?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
maybe it would be easier to multiply both sides by \[x(x-3)\]
anonymous
  • anonymous
i did
anonymous
  • anonymous
but i can't figure out what i did wrong because i get the wrong answer... when i multiply x+2 and x-3 i get x^2+5x+5 when i multiply x-1 and x-3 i get x^2-4x-3

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anonymous
  • anonymous
you get \[(x+2)(x-3)=(x-1)(x-3)-4x+2\] i htink
anonymous
  • anonymous
hold on \[(x+2)(x-3)=x^2-x-6\] ,maybe that is the mistake
anonymous
  • anonymous
also \[(x-1)(x-3)=x^2-4x+3\]
anonymous
  • anonymous
oh!!!!!! i put (x+2) and postive (x+3)!!
anonymous
  • anonymous
that was a waste of time i was looking just at how i solved it and not how i multiplied it. Thanks
anonymous
  • anonymous
\[(x+2)(x-3)=(x-1)(x-3)-4x+2\] \[x^2-x-6=x^2-4x+3-4x+2\] \[x^2-x-6=x^2-8x+5\] etc
anonymous
  • anonymous
yw
anonymous
  • anonymous
when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative? for examples x-1 4x+2 ---- - ----- x(x-3) x(x-3) would you make it into a plus sign, and then make 4x and 2 negative? -4x-2
anonymous
  • anonymous
bump

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